Category Archives: Arithmetic

May 27, 2026 · 12:19 am

Percent

I finally put together some notes (with the help of Copilot) on percent.

Percent Notes Download

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Filed under Arithmetic, Percentages, Year 9

Tagged as amount as a percent, percent, percent decrease, percent increase, percent of an amount, repeated percent

November 11, 2025 · 10:19 am

Completing the Square

$x^2+6x-4 \rightarrow (x+3)^2-13$

Completing the square is useful to

sketch parabolas.
solve quadratics.
factorising quadratics
finding the centre and radius version of the equation of a circle.

When completing the square we take advantage of perfect squares. For example, $(x+3)^2=(x+3)(x+3)=x^2+6x+9$

$6=2\times 3$ and $9=3\times 3$

Example 1

Put $x^2+8x-5$ into completed square form.

What perfect square has an $8x$ term?

$(x+4)^2=x^2+8x+16$

We don’t want $+16$ , we want $-5$ , so subtract $16+5$

$x^2+8x-5=(x+4)^2-21$

$x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c$

What about a non-monic quadratic? For example,

$2x^2+12x+11$

Factorise the $2$

$2(x^2+6x+\frac{11}{2})$

And continue as before

$2[(x+3)^2-9+\frac{11}{2}]=2[(x+3)^2-\frac{18}{2}+\frac{11}{2}]=2[(x+3)^2-\frac{7}{2}]=2(x+3)^2-7$

Example 2

$y=2x^2+7x-5$

$2(x^2+\frac{7}{2}x-\frac{5}{2})$

$2[(x+\frac{7}{4})^2-(\frac{7}{4})^2-\frac{5}{2}]$

$2[(x+\frac{7}{4})^2-\frac{49}{16}-\frac{40}{16}]$

$2[(x+\frac{7}{4})^2-\frac{89}{16}]$

$2(x+\frac{7}{4})^2-\frac{89}{8}$

$ax^2+bx+c=a(x+\frac{b}{2a})^2-a((\frac{b}{2a})^2+\frac{c}{a})$

Casio Classpad e-activity

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Filed under Algebra, Arithmetic, Classpad Skills, Completing the Square, Fractions, Quadratic, Quadratics, Year 10 Mathematics, Year 9 Mathematics

Tagged as completing the square, quadratics

October 4, 2025 · 8:36 am

Converting base 10 numbers to base 2

Converting integers to base 2 is reasonably easy.

For example, what is 82 in base 2?

Think about powers of 2

$n$	$2^n$
$0$	$1$ (‘ones’)
$1$	$2$ (‘tens’)
$2$	$4$ (‘hundreds)
$3$	$8$ (‘thousands’)

Make $82$ the sum of powers of $2$ .

$82=64+16+2=1\times 2^6+0\times 2^5+ 1\times 2^4+0\times 2^3+0\times 2^2+1 \times 2^1+0\times 2^0=1010010$

We follow the same approach for real numbers

$n$	$2^n$
$-3$	$\frac{1}{8}=0.125$
$-2$	$\frac{1}{4}=0.25$
$-1$	$\frac{1}{2}=0.5$
$0$	$1$ (‘ones’)
$1$	$2$ (‘tens’)
$2$	$4$ (‘hundreds)
$3$	$8$ (‘thousands’)

Convert $0.765625$ to base $2$

$0.765625\times 2=1.53125$ the first number is 1

$0.53125\times 2=1.0625$ the second number is 1

$0.0625\times 2=0.125$ the third number is 0

$0.125\times 2=0.25$ the fourth number is 0

$0.25\times 2=0.5$ the fifth number is 0

$0.5\times 2=1$ the sixth number is 1 and we have finished

$0.765625=0.110001_2$

What about something like $2.\overline{4}$ ?

The non-decimal part $2=10$

$0.\overline{4}\times 2=0.\overline{8}$ first number is zero

$0\overline{8}\times 2=1.\overline{7}$ second number is 1

$0.\overline{7}\times 2=1.\overline{5}$ third number is 1

$0.\overline{5}\times 2=1.\overline{1}$ fourth number is 1

$0.\overline{1}\times 2=0.\overline{2}$ fifth number is 0

$0.\overline{2}\times 2=0.\overline{4}$ sixth number is 0

We are back to where we started, so $2.\overline{4}=10.0111000111000..._2=10.\overline{0.11100}_2$

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Filed under Arithmetic, Decimals, Fractions, Number Bases

Tagged as base 2, binary, number bases

September 17, 2025 · 7:03 am

Divisibility Rule for 11

I was working on a question and involved 11 and I wondered what the divisibility rule was?

So then I had a bit of a think about it.

Let $N$ be a number divisible by $11$ . The $N (mod11)=0$

$\begin{equation*}N=a_n\times10^n+a_{n-1}\times10^{n-1}+a_{n-2}\times10^{n-2}+...+a_0\times10^0\end{equation}$

$\begin{equation*}0=a_n\times10^n (mod11)+a_{n-1}\times10^{n-1}(mod11)+a_{n-2}\times10^{n-2}(mod11)+...+a_0\times10^0(mod11)\end{equation}$

Now $10 (mod11)=10$ which is congruent to $-1$ because $10-(-1)=11$ , which is a multiple of 11.

Thus

$\begin{equation*}0=a_n(-1)^n+a_{n-1}(-1)^{n-1}+a_{n-2}(-1)^{n-2}+...+a_0(-1)^0\end{equation}$

Odd powers will be negative and even positive.

So if we start at one end of the number and add every second digit (i.e. first digit plus third digit plus fifth digit etc.) and then subtract the other digits (i.e. second digit, fourth digit, six digit, etc.), if that equals zero then the number is divisible by 11.

For example, is $1756238$ divisible by $11$ ?

$1+5+2+8-7-6-3=16-16=0$

Hence $1756238$ is divisible by $11$

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Filed under Algebra, Arithmetic, Divisibility, Index Laws, Interesting Mathematics, Number Bases

Tagged as Divisibility rules, divisible by 11

July 1, 2025 · 10:35 am

Synthetic Division (for factorising and/or solving polynomials)

I use synthetic division to factorise polynomials with a degree greater than 2. For example, $f(x)=x^3+2x^2-5x-6$

It works best with monic polynomials but can be adapted to non-monic ones (see example below).

The only problem is that you need to find a root to start.

Try the factors of $-6$ i.e. $(-1, 1, 2, -2, 3, -3, 6, -6)$

$f(-1)=(-1)^3+2(-1)^2-5(-1)-6=0$

Hence, $x=-1$ is a root and $(x+1)$ is a factor of the polynomial.

Set up as follows

Bring the first number down

Multiply by the root and place under the second co-efficient

Add down

Repeat the process

The numbers at the bottom $(1, 1, -6)$ are the coefficients of the polynomial factor.

We now know $x^3+2x^2-5x-6=(x+1)(x^2+x-6)$ .

We can factorise the quadratic in the usual way.

$x^2+x-6=(x+3)(x-2)$

Hence $x^3+2x^2-5x-6=(x+1)(x+3)(x-2)$ .

Let’s try a non-monic example

Factorise $6x^4+39x^3+91x^2+89x+30$

I know $-2$ is a root. Otherwise I would try the factors of 30.

Use synthetic division

Because this was non-monic we need to divide our new co-efficients $(6, 27, 37, 15)$ by 6 (the co-efficient of the $x^4$ term)

$x^3+\frac{9}{2}x^2+\frac{37}{6}+\frac{5}{2}$

We now need to go again. I know that $\frac{-3}{2}$ is a root and $(2x+3)$ is a factor.

Our quadratic factor is $x^2+3x+5/3$ , which is $3x^2+9x+5$ .

The quadratic factor doesn’t have integer factors so,

$6x^4+39x^3+91x^2+89x+30=(x+2)(2x+3)(3x^2+9x+5)$

I think this is much quicker than polynomial long division.

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Filed under Algebra, Factorising, Factorising, Fractions, Polynomials, Quadratic, Simplifying fractions, Solving Equations

Tagged as non monic synthetic division, Synthetic division

June 19, 2025 · 7:18 am

Converting recurring (non-terminating) decimals to fractions

The easiest approach is to jump right in with some examples.

Example 1

Convert $0.\overline{5}$ to a fraction.

Let $x=0.\overline{5}$

(1) $\begin{equation*}x=0.\overline{5}\end{equation*}$

(2) $\begin{equation*}10x=5.\overline{5}\end{equation*}$

Subtract equation $1$ from equation $2$

$\begin{equation*}9x=5\end{equation}$

Hence $x=\frac{5}{9}$ so $0.\overline{5}=\frac{5}{9}$

Example 2

Convert $0.\overline{12}$ to a fraction.

Let $x=0.\overline{12}$

(3) $\begin{equation*}x=0.\overline{12}\end{equation*}$

(4) $\begin{equation*}100x=12.\overline{12}\end{equation*}$

Subtract equation $3$ from equation $4$ .

$\begin{equation*}99x=12\end{equation}$

$\begin{equation*}x=\frac{12}{99}=\frac{4}{33}\end{equation}$

Example 3

Convert $0.1\overline{23}$ to a fraction

Let $x=0.1\overline{23}$

(5) $\begin{equation*}x=0.1\overline{23}\end{equation*}$

(6) $\begin{equation*}10x=1.\overline{23}\end{equation*}$

(7) $\begin{equation*}1000x=123.\overline{23}\end{equation*}$

Subtract equation $6$ from equation $7$

$\begin{equation*}990x=122\end{equation}$

$\begin{equation*}x=\frac{122}{990}=\frac{61}{495}\end{equation}$

Our aim is to manipulate the recurring decimal to create two numbers each which have only the repeated digits after the decimal point.

One more example.

Example 4

Convert $3.4\overline{56}$ to a fraction

Let $x=3.4\overline{56}$

If I multiply by 10, then I will have $34.\overline{56}$ – only repeated digits after the decimal point.

If I multiply by 1000, then I will have $3456.\overline{56}$ – only repeated digits after the decimal point.

So I get,

$\begin{equation*}990x=3422\end{equation}$

$\begin{equation*}x=\frac{3422}{990}=3\frac{226}{495}\end{equation}$

You can also use your Casio classpad to do the conversion. Although I think it is easier just to do it yourself.

Let’s think about example 4,

$3.4\overline{56}=3.4+\frac{56}{1000}+\frac{56}{100000}+\frac{56}{10000000}+...$

Which is

$3.4+\frac{56}{1000\times 100^0}+\frac{56}{1000\times100^1}+\frac{56}{1000\times 100^2}...$

$3.4+\Sigma_{x=0}^\infty(\frac{56}{1000\times 100^x})$

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Filed under Arithmetic, Decimals, Fractions, Year 11 Specialist Mathematics

Tagged as recuring decimals to fractions, Year 11 Specialist Mathematics

May 15, 2025 · 10:01 am

Perfect Squares

Find all of the positive integers that make the following expression a perfect square.

(1) $\begin{equation*}(x-10)(x+14)\end{equation*}$

Let

$\begin{equation*}(x-10)(x+14)=n^2\end{equation}$

where $n$ is an integer.

Expand and simplify

$\begin{equation*}x^2+4x-140=n^2\end{equation}$

$\begin{equation*}x^2-4x-n^2=140\end{equation}$

Complete the square

$\begin{equation*}(x+2)^2-4-n^2=140\end{equation}$

$\begin{equation*}(x+2)^2-n^2=144\end{equation}$

Factorise (using difference of perfect squares)

$\begin{equation*}(x+2-n)(x+2+n)=144\end{equation}$

Find all of the factors of $144$

$(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12)$

First pair,

$\begin{equation*}x+2-n=1 \tag {1} \end{equation}$

$\begin{equation*}x+2+n=144 \tag {2} \end{equation}$

$2x=141$

$x$ must be an integer.

I then used a spreadsheet

Hence the integers that make $(x-10)(x+14)$ are perfect square are, $10, 11, 13, 18,$ and $35$ .

Let’s try another one,

$(x-6)(x+14)$

(2) $\begin{equation*}(x-6)(x+14)=n^2\end{equation*}$

$\begin{equation*}(x^2+8x-84=n^2\end{equation}$

$\begin{equation*}(x+4)^2-n^2=100\end{equation}$

$\begin{equation*}(x+4-n)(x+4+n)=100\end{equation}$

Factors of 100,

$(1, 100), (2, 50), (4, 25), (5, 20), (10, 10)$

So the possible integers are $6$ and $22$ .

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Filed under Algebra, Arithmetic, Divisibility, Interesting Mathematics, Puzzles, Quadratic, Solving Equations

Tagged as perfect squares

September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let $x$ be the number of cats and $y$ be the number of dogs.
Then $0.9x+0.1y$ think they are cats.
But we also know 20% of the total think they are cats.
$0.2(x+y)$
Therefore, $0.9x+0.1y=0.2(x+y)$
$0.9x+0.1y=0.2x+0.2y$
$0.7x=0.1y$
$7x=y$
Percentage of cats is $\frac{x}{x+y}\times100$
Substitute $7x$ for $y$
$\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%$
$\therefore 12.5$ % of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

Tagged as Fractions, percent, uk maths challenge

July 16, 2024 · 7:35 am

Australian Mathematics Competition – Polynomial Question

I came across this question from the 2010 Senior Australian Mathematics Competition:

A polynomial $f$ is given. All we know about it is that all its coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$ . What is the value of $f(3)$
Australian Mathematics Competition 2006-2012

I thought ‘excellent, a somewhat hard polynomial question for my students’ and then I tried it. Now I know why only 1% of students got it correct.

As we don’t know the order of the polynomial, let

$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0$

We know all of the coefficients are greater than or equal to zero. We also know

$f(1)=a_n+a_{n-1}+...+a+a_0=6$

Which means that all of the coefficients are between zero and six

$0\le{a_n}\le{6}$

We have also been given $f(7)$

$f(7)=7^na_n+7^{n-1}a_{n-1}+ ... +7a+1=3438$

As all of the coefficients are between zero and six, this is $3438$ written in base 7.

Let’s calculate a few powers of 7

	Powers of 7
$7^0$	1
$7^1$	7
$7^2$	49
$7^3$	343
$7^4$	2401
$7^5$	16807

Hence $3438$ written in base $7$ is $13011$

Therefore $f(x)=x^4+3x^3+x+1$

$f(3)=3^4+3\times3^3+3+1$

$f(3)=81+81+4$

$f(3)=166$

I really like this question. I think it could work well as a class extension activity with a bit of scaffolding.

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Filed under Number Bases, Polynomials

Tagged as base 7, number bases, Polynomials

May 12, 2024 · 11:50 am

Fractions to decimals

People usually know some fractions as decimals, for example

$\frac{1}{4}=0.25\ \textnormal{or }\frac{4}{5}=0.8$

And denominators that are powers of ten are also easy,

$\frac{47}{100}=0.47\ \textnormal{or }\frac{256}{1000}=0.256$

But what if it is something else? One that you don’t know. For example,

$\frac{5}{12}\ \textnormal{or }\frac{15}{37}$

I like to do these as a long division

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Filed under Arithmetic, Decimals, Fractions

Tagged as decimals, Fractions, fractions to decimals

As numbers	As Powers of 7
$3438=1\times2401+1037$	$3438=1\times7^4+1037$
$1037=3\times343+8$	$1037=3\times7^3+8$
$8=1\times7+1$	$8=1\times7^1+1$
$1=1\times1$	$1=1\times7^0$

Category Archives: Arithmetic

Percent

Completing the Square

Example 1

Example 2

Converting base 10 numbers to base 2

Divisibility Rule for 11

Synthetic Division (for factorising and/or solving polynomials)

Converting recurring (non-terminating) decimals to fractions

Perfect Squares

Cats and Dogs

Australian Mathematics Competition – Polynomial Question

Fractions to decimals

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